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Using Newton's Method, we wish to find a root of:

$$f(x)=\sin(x)-\tan(x)+1$$

...starting with $x_0=\dfrac{\pi}{4}$. Find $x_1$.

A

$\dfrac{\pi}{4}-\dfrac{1}{2}$

B

$\dfrac{\pi}{4}+\dfrac{\sqrt{2}-4}{\sqrt{2}}$

C

$\dfrac{\pi}{2}-\dfrac{\sqrt{2}}{\sqrt{2}-2}$

D

$\dfrac{\pi}{4}-\dfrac{\sqrt{2}}{\sqrt{2}-4}$

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